Key Concepts and Formulas

• Limit Evaluation Techniques: Indeterminate forms ($0/0$, $\infty/\infty$), L'Hôpital's Rule, standard limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$, and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.
• Limits of the form $1^\infty$: If $\lim_{x \to c} f(x) = 1$ and $\lim_{x \to c} g(x) = \infty$, then $\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} g(x) [f(x) - 1]}$. Alternatively, $\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} g(x) \ln(f(x))}$.
• Quadratic Equations: If $\alpha$ and $\beta$ are the roots of a quadratic equation $ax^2 + bx + c = 0$, then the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.

Step-by-Step Solution

Part 1: Evaluate $\alpha$

Step 1: Analyze the limit for $\alpha$. The given limit for $\alpha$ is: $\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ As $x \to \frac{\pi}{4}$, $\tan x \to \tan(\frac{\pi}{4}) = 1$. So, the numerator approaches $1^3 - 1 = 0$. As $x \to \frac{\pi}{4}$, $x + \frac{\pi}{4} \to \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. So, $\cos(x + \frac{\pi}{4}) \to \cos(\frac{\pi}{2}) = 0$. This is an indeterminate form of type $\frac{0}{0}$.

Step 2: Apply L'Hôpital's Rule to evaluate $\alpha$. Since we have the $\frac{0}{0}$ indeterminate form, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$. Derivative of the numerator: $\frac{d}{dx}(\tan^3 x - \tan x) = 3\tan^2 x \cdot \sec^2 x - \sec^2 x$ Derivative of the denominator: $\frac{d}{dx}\left(\cos \left( {x + {\pi \over 4}} \right)\right) = -\sin \left( {x + {\pi \over 4}} \right) \cdot \frac{d}{dx}\left(x + \frac{\pi}{4}\right) = -\sin \left( {x + {\pi \over 4}} \right) \cdot 1$ Applying L'Hôpital's Rule: $\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} \frac{3\tan^2 x \sec^2 x - \sec^2 x}{-\sin \left( {x + {\pi \over 4}} \right)}$

Step 3: Substitute the value of $x$ after applying L'Hôpital's Rule. Now, substitute $x = \frac{\pi}{4}$: $\tan(\frac{\pi}{4}) = 1$ $\sec(\frac{\pi}{4}) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$, so $\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$. $x + \frac{\pi}{4} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. $\sin(\frac{\pi}{2}) = 1$. Plugging these values into the limit expression: $\alpha = \frac{3(1)^2 (2) - 2}{-\sin \left( {\frac{\pi}{2}} \right)} = \frac{3(2) - 2}{-1} = \frac{6 - 2}{-1} = \frac{4}{-1} = -4$ So, $\alpha = -4$.

Part 2: Evaluate $\beta$

Step 4: Analyze the limit for $\beta$. The given limit for $\beta$ is: $\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$ As $x \to 0$, $\cos x \to \cos(0) = 1$. As $x \to 0$, $\cot x = \frac{\cos x}{\sin x} \to \frac{1}{0^+}$, which approaches $\infty$. This is an indeterminate form of type $1^\infty$.

Step 5: Use the standard form for $1^\infty$ limits to evaluate $\beta$. We use the formula: If $\lim_{x \to c} f(x) = 1$ and $\lim_{x \to c} g(x) = \infty$, then $\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} g(x) [f(x) - 1]}$. Here, $f(x) = \cos x$ and $g(x) = \cot x$. $\beta = e^{\mathop {\lim }\limits_{x \to 0} \cot x (\cos x - 1)}$

Step 6: Evaluate the exponent's limit. Let's evaluate the limit in the exponent: $L = \mathop {\lim }\limits_{x \to 0} \cot x (\cos x - 1) = \mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\sin x} (\cos x - 1)$ We can rewrite this as: $L = \mathop {\lim }\limits_{x \to 0} \frac{\cos x - 1}{\sin x} \cdot \cos x$ As $x \to 0$, $\cos x \to 1$, so the limit becomes: $L = \mathop {\lim }\limits_{x \to 0} \frac{\cos x - 1}{\sin x} \cdot 1 = \mathop {\lim }\limits_{x \to 0} \frac{\cos x - 1}{\sin x}$ This is still a $\frac{0}{0}$ indeterminate form. We can use L'Hôpital's Rule or manipulate it using standard limits. Let's use standard limits for practice. We know $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. $L = \mathop {\lim }\limits_{x \to 0} \frac{-(1 - \cos x)}{\sin x}$ Multiply and divide by $x^2$ and $x$ respectively to use the standard limits: $L = \mathop {\lim }\limits_{x \to 0} \frac{-(1 - \cos x)}{x^2} \cdot \frac{x^2}{\sin x}$ $L = \mathop {\lim }\limits_{x \to 0} \left( -\frac{1 - \cos x}{x^2} \right) \cdot \left( \frac{x}{\sin x} \cdot x \right)$ $L = \left( -\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos x}{x^2} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0} x \right)$ Using the standard limits: $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\lim_{x \to 0} \frac{x}{\sin x} = 1$. $L = \left( -\frac{1}{2} \right) \cdot (1) \cdot (0) = 0$

Alternatively, using L'Hôpital's rule on $\mathop {\lim }\limits_{x \to 0} \frac{\cos x - 1}{\sin x}$: $L = \mathop {\lim }\limits_{x \to 0} \frac{-\sin x}{\cos x} = \frac{-\sin 0}{\cos 0} = \frac{0}{1} = 0$ So, the limit of the exponent is 0.

Step 7: Determine the value of $\beta$. Since the limit of the exponent is 0, we have: $\beta = e^0 = 1$ So, $\beta = 1$.

Part 3: Use the roots to find a and b

Step 8: Set up the quadratic equation using the roots $\alpha$ and $\beta$. We found $\alpha = -4$ and $\beta = 1$. These are the roots of the equation $ax^2 + bx - 4 = 0$. For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $-\frac{B}{A}$ and the product of roots is $\frac{C}{A}$. In our equation $ax^2 + bx - 4 = 0$, we have $A=a$, $B=b$, and $C=-4$.

Step 9: Use the sum and product of roots to form equations for a and b. Sum of roots: $\alpha + \beta = -4 + 1 = -3$. From the equation, the sum of roots is $-\frac{b}{a}$. So, $-3 = -\frac{b}{a} \implies \frac{b}{a} = 3 \implies b = 3a$. (Equation 1)

Product of roots: $\alpha \beta = (-4)(1) = -4$. From the equation, the product of roots is $\frac{-4}{a}$. So, $-4 = \frac{-4}{a}$. This implies $a = 1$.

Step 10: Solve for b using the value of a. Substitute $a=1$ into Equation 1: $b = 3a = 3(1) = 3$.

Thus, $a = 1$ and $b = 3$. The ordered pair is $(a, b) = (1, 3)$.

Step 11: Verify the result by substituting a and b back into the quadratic equation. The equation is $ax^2 + bx - 4 = 0$. With $a=1$ and $b=3$, the equation becomes $1x^2 + 3x - 4 = 0$, or $x^2 + 3x - 4 = 0$. We can factor this equation: $(x+4)(x-1) = 0$. The roots are $x = -4$ and $x = 1$. These are indeed our calculated values for $\alpha = -4$ and $\beta = 1$.

Common Mistakes & Tips

• Incorrect application of L'Hôpital's Rule: Ensure that the limit is in an indeterminate form ($0/0$ or $\infty/\infty$) before applying L'Hôpital's Rule. Differentiate the numerator and denominator separately.
• Algebraic errors in limit evaluation: Be careful with signs and cancellations when manipulating expressions, especially when using standard limits.
• Misinterpreting the $1^\infty$ form: The formula $e^{\lim g(x)(f(x)-1)}$ is crucial for such limits. Do not directly substitute values when the form is $1^\infty$.
• Relating roots to coefficients: Remember that for $ax^2+bx+c=0$, sum of roots is $-b/a$ and product of roots is $c/a$. In this problem, the constant term is -4, which is directly related to the product of roots.

Summary

The problem requires us to first evaluate two limits, $\alpha$ and $\beta$. For $\alpha$, we encountered a $\frac{0}{0}$ indeterminate form, which we resolved using L'Hôpital's Rule to find $\alpha = -4$. For $\beta$, we faced a $1^\infty$ indeterminate form, which we converted to an exponential form using the standard limit formula $e^{\lim g(x)(f(x)-1)}$, and then evaluated the exponent's limit using standard trigonometric limits or L'Hôpital's Rule to find $\beta = 1$. Once the roots $\alpha = -4$ and $\beta = 1$ were determined, we used Vieta's formulas for the quadratic equation $ax^2 + bx - 4 = 0$. The product of roots $\alpha\beta = \frac{-4}{a}$ gave us $a=1$, and the sum of roots $\alpha+\beta = -\frac{b}{a}$ yielded $b=3$. Therefore, the ordered pair $(a, b)$ is $(1, 3)$.

The final answer is \boxed{(1, 3)}.